Question

How many grams of dry NH4Cl need to be added to 1.60 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.81? Kb for ammonia is 1.8×10−5.

Answer #1

NH4Cl ----------> NH4(+) + Cl(-)

NH3 + H2O <----------> NH4(+) + OH(-)

The Kb is small so the it is possible to say that all the NH4(+) is
released from NH4Cl and the concentration of NH3 will not
change

pOH = 14-pH

pOH = 14- 8.81

pOH = 5.19

from this concentration of OH-

[OH-] = 10^{-pOH}

[OH-] = 10^{-5.19}

[OH-] = 6.46 x 10^{-6}

Kb = [OH-][NH4+] / [NH3]

1.8×10^{−5} = [6.46 x 10^{-6}][NH4+] / 0.8

[NH4+]= 1.44 x 10^{-5} / 6.46 x 10^{-6}

[NH4+] = 2.23M

from molarity we can calculate the molarity

2.23 = = (weight of NH4Cl / molar mass of NH4Cl) x 1/V in liters

weight of NH4 Cl = 2.33 M x 1.6 L x 53.5

= 199.45 grams

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